Question

A hot black painted pipe(2.54-cmO.D.and50-mlong)passesthrougha 20°C room to heat it. The temperature of the pipe surface is 150°C. If the heat transfer coefficient by convection to the air is 10 W/m2-K :

a) What is the total heat rate loss?

b) What is the radiation heat transfer coefficient?

c) At what temperature of the pipe, the heat rate by convection equals the heat rate by radiation.

Answer 1

Given data

Dia D = 2.54 cm x 1m/100cm = 0.0254 m

Length L = 50 m

Area A = 3.14 x D x L = 3.14 x 0.0254 x 50 = 3.9878 m2

Tsurr = 20 C

Pipe surface temperature Ts = 150 C

Heat transfer coefficient h = 10 W/m2-K

Part a

Total heat rate loss

Qconv = h x A x (Ts - Tsurr)

= 10 x 3.9878 x (150 - 20)

= 5184.14 W

Part b

Radiation heat transfer

Qrad = A (Ts⁴ - Tsurr⁴) = hr x A x (Ts - Tsurr)

radiation heat transfer coefficient

hr = (Ts⁴ - Tsurr⁴) / (Ts - Tsurr)

= 5.67 x 10^-8 W/m2-K4 x (150⁴ - 20⁴)K4/(150 - 20)K

= 28.695303 / 130

= 0.2207 W/m2-K

Part c

Qconv = Qrad

h x (Ts - Tsurr) = (Ts⁴ - Tsurr⁴)

10 x (Ts - 20) = 5.67 x 10^-8 x (Ts⁴ - 20⁴)

Ts - 20 = 5.67 x 10^-9 x Ts⁴ - 0.0009072

Ts - 5.67 x 10^-9 x Ts⁴ - 20 + 0.0009072 = 0

Ts = 553 K